Nimber multiplication In the question on nimbers, the original poster asks for the meaning of Nimber multiplication in the context of impartial games.

Edit: As noted by Mark Fischler in the comments below, the following is wrong
My gut instinct is $*a \times *b$ means that if $*a$ is a game equivalent to $a$ stones, and $*b$ is a game equivalent to $b$ stones, then if you replace every stone in the $*a$ game with a copy of the $*b$ game, you get a game with the Nimber $*a \times *b$, but I haven't been able prove it.
 A: I don't think there's an intuitive way to understand nimber multiplication. Also note that it is distinct from repeated addition, so it isn't as simple as replacing stones with copies of piles. Multiplication is defined recursively as
$$ab = \text{mex}\left(\{a'b+ab'+a'b': a'<a, b'<b\}\right)$$
where $\text{mex}$ is the minimum excluded element.
Although it's hard to see what this means in terms of a concrete game, it's possible to understand why it's defined this way. The point is that we want to create an algebraic system without zero divisors, so $(a-a')(b-b') \neq 0$ whenever $a\neq a'$ and $b\neq b'$. In other words,
$$ab - ab' - a'b + a'b' \neq 0,$$
so
$$ab \neq a'b + ab' - a'b'.$$
Since subtraction and addition are the same operation, we have
$$ab \neq a'b + ab' + a'b'.$$
The definition above takes the first nimber that meets this criterion.
A: Winning Ways Volume 3, Chapter 14, considers coin-turning games. One such game is Turning Corners, where a move is to turn over the four corners of a rectangle with horizontal and vertical sides, but the North-East most coin needs to be turned from heads to tails. A sample move shows that it defines nimber multiplication:

a           X           H
.  
.  
.  
a'          X           X
.  
.  
.  
0  .  .  .  b' .  .  .  b

A move will turn the H to tails and Xs over, so from the position $(a,b)$, a move moves to the sum $(a',b)+(a,b')+(a',b')$, so the value of the position $(a,b)$ is the nimber product $ab$ inductively.
A: The "game" in this video may be analyzed in terms of a limited form of nimber multiplication, specifically involving the nimbers $1,2,3$.
The rules of the game are as follows:

*

*Start with disks (or hexagons in the video) of three different colors. Fill a base row of a triangular array with some permutation of $n$ disks.


*This generates $n-1$ adjacent pairs. On top of each adjacent pair add a disk of the same color if the two disks below are identical, add a disk of the third color otherwise. Repeat this process for the next layer up until you reach one final disk at the top.
For certain values of the layer count $n$ the color of the top disk will depend only on the two other corners of the triangle; all the other disks in the base layer and all the intermediate layers cancel themselves out. The special layer counts are
$n=2,4,10,...; n\in\{3^k+1,k\in\mathbb{N}\}.$
We may analyze this in terms of nimber multiplication by rendering the colors as the nimbers $1,2,3$. With this rendering the placement rule is simply that the nimber product of any two horizontally adjacent disks with the one above is $1$, to wit
$1\otimes1\otimes1=2\otimes2\otimes2=3\otimes3\otimes3=1\otimes2\otimes3=1.$
Thus disks with colors $a$ and $b$ will be topped by a disk whose color corresponds to $a^{-1}b^{-1}$, where the product is understood to be the nimber product. For an $n$-layer array, the top disk will have a combined product of all the base disks raised to some powers; for instance with six layers the top disk carries the product
$a^{-1}b^{-5}c^{-10}d^{-10}e^{-5}f^{-1}$
where the disks are in order from $a$ to $f$ across the base row. The exponent will in general be $(-1)^{n-1}$ times the entry in Pascal's Triangle corresponding to the disk's position. The exponents always sum to $(-2)^{n-1}\equiv1\bmod 3$.
Since each nimber cubed is $1$, the product may be simplified by reducing each exponent to a residue $\in\{-1,0,1\}\bmod 3$. For the six-layer case above this gives
$a^{-1}bc^{-1}d^{-1}ef^{-1}$
whereas a product depending only on the corners would have been $a^{-1}f^{-1}$. Thus six layers is not a "special number". But for four layers we do properly get $a^{-1}d^{-1}$ as the intervening elements in Pascal's Triangle are multiples of $3$ ($1~\color{blue}{3~3}~1$). We see, given well-known properties of Pascal's Triangle, the "special numbers" correspond to $n-1$ being a power of $3$ as indicated in the video.
